## #980 integrate(1/(sin(x)^2+1),x,1,1+%pi) takes forever

closed
nobody
5
2007-06-22
2006-09-05
Raymond Toy
No

This is a follow-on to Bug [ 1044318 ]
defint(1/(sin(x)^2+1),x,0,3*%pi) wrong.

integrate(1/(sin(x)^2+1),x,1,1+%pi) seems to take
forever. But if the limits are 0 and %pi, the integral
is evaluated instantly with the correct value.

## Discussion

• Barton Willis - 2006-09-06

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ldefint gives a wrong answer (basically bug 1044318)

(%i10) ldefint(1/(sin(x)^2+1),x,1,1+%pi);
(%o10) 0

integrate gives an antiderivative that isn't continuous
at odd integer multiples of %pi / 2

(%i11) integrate(1/(sin(x)^2+1),x);
(%o11) atan((2*tan(x))/sqrt(2))/sqrt(2)

The expression
atan(2*tan(x)/sqrt(2))/sqrt(2)+sqrt(2)*%pi*floor(x/%pi-1/2)/2

might be a valid antiderivative on all of R provided the
first term is assumed left continuous at each odd integer
multiple of %pi/2, I think.

Barton

• Raymond Toy - 2006-09-06

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Looking at the code for ldefint shows why it returns 0. As
the docs say, it evaluates the antiderivative and plugs in
the limits.

I believe this is why defint uses the routine intsubs and
same-sheet-subs to handle these issues. Perhaps they're not
doing what they're supposed to do, but in this particular
case, maxima is stuck computing the antiderivative. I don't
understand why.

• Raymond Toy - 2007-06-22

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This integral no longer takes forever. Don't know how it got fixed. The result is wrong, though. See bug 1741705

Closing this since it no longer directly applies.

• Raymond Toy - 2007-06-22
• status: open --> closed